This is more of a long comment than an answer. It should be possible
to compute the number of regions and number of bounded regions using
Whitney's theorem for the characteristic polynomial $\chi(t)$ (Theorem
2.4 of <a
href="https://www.cis.upenn.edu/~cis610/sp06stanley.pdf">these
notes</a>), and Zaslavsky's theorem that the number of regions is
$(-1)^d \chi(-1)$, and the number of bounded regions is (in this
situation) $(-1)^d\chi(1)$ (Theorem 2.5 of the previous link). We need more
than the usual definition of "general position." We want the
position to be generic enough for the argument below (generalized to
$d$ dimensions) to hold.

Here is the computation for $d=2$. First, the empty intersection (the
ambient space $\mathbb{R}^2$) contributes $t^2$ to $\chi(t)$. The
${n\choose 2}$ lines will contribute $-{n\choose 2}t$. Now we must
consider all subsets of the lines that intersect in a point $p$. Let
$p$ be one of the original $n$ points. Then ${n-1\choose 2}$ pairs of
lines intersect in $p$, ${n-1\choose 3}$ triple of lines intersect in
$p$, etc., giving a contribution to $\chi(t)$ of
  $$ {n-1\choose 2}-{n-1\choose 3}+{n-1\choose 4} -\cdots = n-2. $$
We have to multiply this by $n$ since there are $n$ choices for
$p$. There are now $3{n\choose 4}$ choices of two lines that don't
intersect in one of the original $n$ points, but they still intersect
by genericity. Thus we get an additional contribution of $3{n\choose
4}$. It follows that
  $$ \chi(t) = t^2-{n\choose 2}t+n(n-2)+3{n\choose 4}. $$
The number of regions is
  $$ \chi(-1) = \frac 18(n-1)(n^3-5n^2+18n-8). $$
The number of bounded regions is
  $$ \chi(1) = \frac 18(n-1)(n-2)(n^2-3n+4). $$
Can someone extend this argument to $d$ dimensions?

**Addendum.** I worked out $d=3$. Here are the details. Let $X$ be an
  $n$-element "generic" subset of $\mathbb{R}^3$. Thus $X$ determines
  a set $\mathcal{A}$ of ${n\choose 3}$ hyperplanes. We need to find
  all subsets of $\mathcal{A}$ with nonempty intersection. A
  $j$-element subset that intersects in an $e$-dimensional affine
  space contributes $(-1)^jt^e$ to the characteristic polynomial
  $\chi(t)$.

  *Case 1:* $e=3$. We take the intersection over the empty set to get
   $\mathbb{R}^3$. This gives a term $t^3$.

  *Case 2:* $e=2$. Each hyperplane contributes $-t^2$, giving a term
    $-{n\choose 3}t^2$.

  *Case 3:* $e=1$. (a) Any two hyperplanes intersect in a line (by
  genericity), giving ${{n\choose 3}\choose 2}t$.

   (b) Any $j\geq 3$ hyperplanes containing the same two points
   $p,q\in X$ meet in a line. There are ${n\choose 2}$ choices for
   $p,q$ and ${n-2\choose j}$ for the remaining element of $X$ in the
   hyperplanes. Thus we get a contribution
     $$ {n\choose 2}\sum_{j\geq 3}(-1)^j {n-2\choose j}t =
        {n\choose 2}\left[-1+(n-2)-{n-2\choose 3}\right]t. $$

   *Case 4:* $e=0$. (a) Any three hyperplanes intersect at a point,
   except when all three contain the same two points $p,q\in X$. There
   are ${{n\choose 3}\choose 3}$ ways to choose three hyperplanes, and
   ${n\choose 2}{n-2\choose 3}$ ways to choose them so that they
   intersect in two points of $X$. Hence we get a contribution
     $$ -\left[ {{n\choose 3}\choose 3}-{n\choose 2}{n-2\choose 3}
         \right] $$
   to the constant term of $\chi(t)$ (the minus sign because the
   number of hyperplanes is odd).

  (b) Any $j\geq 4$ hyperplanes meeting at a point $p\in X$. We can
  choose $p$ in $n$ ways. We then must choose $j$ two-element subsets
  of $X-p$ whose intersection is empty. There are ${{n-1\choose
  2}\choose j}$ ways to choose $j$ two-element subsets of $X-p$. If
  their intersection is nonempty, then they have a common element $q$
  which can be chosen in $n-1$ ways, and then we can choose the
  remaining elements in ${n-2\choose j}$ ways. This gives the
  contribution
   $$ n\sum_{j\geq 4}(-1)^j\left[ {{n-1\choose 2}\choose j}-
      (n-1){n-2\choose j}\right] $$
   $$ \ = n\left[ -1+{n-1\choose 2}-{{n-1\choose 2}\choose 2} +
       {{n-1\choose 2}\choose 3}-(n-1)\left(-1+(n-2)
       -{n-2\choose 2}+{n-2\choose 3}\right)\right]. $$

   (c) Any $j\geq 3$ hyperplanes meeting at $p,q\in X$, together with
   one additional hyperplane not containing $p$ or $q$. There are
   ${n\choose 2}$ choices for $p,q$ and ${n-2\choose j}$ ways to
   choose $j$ hyperplanes containing $p,q$. There are then
   ${n-2\choose 3}$ ways to choose the additional hyperplane not
   containing $p$ or $q$. Thus we get the contribution
     $$ -\left[ \sum_{j\geq 3}(-1)^j{n-2\choose j}\right]
         {n-2\choose 3} $$
     $$ -{n-2\choose 3}{n\choose 2}\left[-1+(n-2)-{n-2\choose    2}  
        \right]. $$

   Putting all this together gives the characteristic polynomial
     $$ t^3-{n\choose 3}t^2+
        \frac{1}{72}n(n-1)(n-3)(n^3-2n^2-16n+68)t $$
     $$ -\frac{1}{1296}n(n-2)(n-3)(n^6-4n^5-74n^4+698n^3-2129n^2
        +2276n-120). $$
  The number of regions is
   $$ \frac{1}{1296}(n-2)(n^8-7n^7-62n^6+938n^5-4295n^4+8429n^3
      -4932n^2-2016n-648). $$
  The number of bounded regions is
  $$ \frac{1}{1296}(n-1)(n-2)(n-3)(n^6-3n^5-77n^4+603n^3
      -1508n^2+1056n+216). $$
 Conceivably there could be an error in the computation, but I checked
 it for $n=4,5$ by a brute force computation.

 This method should extend to any $d$, but the computation will be
 more complicated, and I am too lazy to work out the details.